Anyone able to work out how many unique solutions there are to the Go First dice problem? http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/sep/18...
Dunno --- but it must be either one or zero! WFL On 9/18/12, Alex Bellos <alexanderbellos@googlemail.com> wrote:
Anyone able to work out how many unique solutions there are to the Go First dice problem?
http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/sep/18... _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Neat idea! Synopsis for people who didn't click through: a set of four 12-sided dice, faces numbered from 1 to 48 with each number appearing once, such that if you roll all four of them, all four permutations of the dice are equally likely. If you only want to do this for two players, then a coin (2-sided die) suffices: number the coins {1,4} and {2,3}. With three people, it looks to me like you can do it with six-sided dice. Here are the 11 solutions, according to a little Mma, though I'd be happy to have someone else confirm that I didn't mess things up: {{1, 5, 9, 12, 14, 16}, {2, 6, 7, 11, 13, 18}, {3, 4, 8, 10, 15, 17}}, {{1, 5, 10, 11, 13, 17}, {2, 6, 8, 9, 14, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 8, 11, 14, 17}, {2, 5, 7, 12, 15, 16}, {3, 4, 9, 10, 13, 18}}, {{1, 6, 8, 11, 14, 17}, {2, 5, 9, 10, 13, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 8, 11, 15, 16}, {2, 5, 7, 12, 14, 17}, {3, 4, 9, 10, 13, 18}}, {{1, 6, 8, 11, 15, 16}, {2, 5, 9, 10, 13, 18}, {3, 4, 7, 12, 14, 17}}, {{1, 6, 8, 12, 13, 17}, {2, 4, 9, 11, 15, 16}, {3, 5, 7, 10, 14, 18}}, {{1, 6, 9, 10, 14, 17}, {2, 5, 7, 12, 15, 16}, {3, 4, 8, 11, 13, 18}}, {{1, 6, 9, 10, 14, 17}, {2, 5, 8, 11, 13, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 9, 10, 15, 16}, {2, 5, 7, 12, 14, 17}, {3, 4, 8, 11, 13, 18}}, {{1, 6, 9, 10, 15, 16}, {2, 5, 8, 11, 13, 18}, {3, 4, 7, 12, 14, 17}}} Smaller than six sides won't work, because the d^3 possible outcomes from rolling three d-sided dice must be divisible by 6. The linked article claims that you can't do four players with six-sided dice (even though 24 | 6^4), and is asking for the number of solutions with 12-sided ones. --Michael On Tue, Sep 18, 2012 at 7:25 AM, Alex Bellos <alexanderbellos@googlemail.com
wrote:
Anyone able to work out how many unique solutions there are to the Go First dice problem?
http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/sep/18... _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
If we relax the constraint that all the dice have the same number of sides, then the two-person problem can be solved with only three "faces". One person "rolls" a device that always yields 2; the other has a coin with 1 and 3 on its faces. Does the three-person problem really need 18 faces? On Thu, Sep 20, 2012 at 10:57 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
Neat idea! Synopsis for people who didn't click through: a set of four 12-sided dice, faces numbered from 1 to 48 with each number appearing once, such that if you roll all four of them, all four permutations of the dice are equally likely.
If you only want to do this for two players, then a coin (2-sided die) suffices: number the coins {1,4} and {2,3}.
With three people, it looks to me like you can do it with six-sided dice. Here are the 11 solutions, according to a little Mma, though I'd be happy to have someone else confirm that I didn't mess things up:
{{1, 5, 9, 12, 14, 16}, {2, 6, 7, 11, 13, 18}, {3, 4, 8, 10, 15, 17}}, {{1, 5, 10, 11, 13, 17}, {2, 6, 8, 9, 14, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 8, 11, 14, 17}, {2, 5, 7, 12, 15, 16}, {3, 4, 9, 10, 13, 18}}, {{1, 6, 8, 11, 14, 17}, {2, 5, 9, 10, 13, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 8, 11, 15, 16}, {2, 5, 7, 12, 14, 17}, {3, 4, 9, 10, 13, 18}}, {{1, 6, 8, 11, 15, 16}, {2, 5, 9, 10, 13, 18}, {3, 4, 7, 12, 14, 17}}, {{1, 6, 8, 12, 13, 17}, {2, 4, 9, 11, 15, 16}, {3, 5, 7, 10, 14, 18}}, {{1, 6, 9, 10, 14, 17}, {2, 5, 7, 12, 15, 16}, {3, 4, 8, 11, 13, 18}}, {{1, 6, 9, 10, 14, 17}, {2, 5, 8, 11, 13, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 9, 10, 15, 16}, {2, 5, 7, 12, 14, 17}, {3, 4, 8, 11, 13, 18}}, {{1, 6, 9, 10, 15, 16}, {2, 5, 8, 11, 13, 18}, {3, 4, 7, 12, 14, 17}}}
Smaller than six sides won't work, because the d^3 possible outcomes from rolling three d-sided dice must be divisible by 6.
The linked article claims that you can't do four players with six-sided dice (even though 24 | 6^4), and is asking for the number of solutions with 12-sided ones.
--Michael
On Tue, Sep 18, 2012 at 7:25 AM, Alex Bellos < alexanderbellos@googlemail.com
wrote:
Anyone able to work out how many unique solutions there are to the Go First dice problem?
http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/sep/18...
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Thu, Sep 20, 2012 at 11:30 AM, Allan Wechsler <acwacw@gmail.com> wrote:
If we relax the constraint that all the dice have the same number of sides, then the two-person problem can be solved with only three "faces". One person "rolls" a device that always yields 2; the other has a coin with 1 and 3 on its faces. Does the three-person problem really need 18 faces?
Great question. If it could be done with all dice having #faces in {1,2,3,6}, then it could be expressed as a 3d6 solution (with appropriately-clustered numbers), and if my computation of the eleven 3d6 solutions is right, then none of them fit the bill. But this doesn't rule out e.g. something using a d4. --Michael
On Thu, Sep 20, 2012 at 10:57 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
Neat idea! Synopsis for people who didn't click through: a set of four 12-sided dice, faces numbered from 1 to 48 with each number appearing once, such that if you roll all four of them, all four permutations of the dice are equally likely.
If you only want to do this for two players, then a coin (2-sided die) suffices: number the coins {1,4} and {2,3}.
With three people, it looks to me like you can do it with six-sided dice. Here are the 11 solutions, according to a little Mma, though I'd be happy to have someone else confirm that I didn't mess things up:
{{1, 5, 9, 12, 14, 16}, {2, 6, 7, 11, 13, 18}, {3, 4, 8, 10, 15, 17}}, {{1, 5, 10, 11, 13, 17}, {2, 6, 8, 9, 14, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 8, 11, 14, 17}, {2, 5, 7, 12, 15, 16}, {3, 4, 9, 10, 13, 18}}, {{1, 6, 8, 11, 14, 17}, {2, 5, 9, 10, 13, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 8, 11, 15, 16}, {2, 5, 7, 12, 14, 17}, {3, 4, 9, 10, 13, 18}}, {{1, 6, 8, 11, 15, 16}, {2, 5, 9, 10, 13, 18}, {3, 4, 7, 12, 14, 17}}, {{1, 6, 8, 12, 13, 17}, {2, 4, 9, 11, 15, 16}, {3, 5, 7, 10, 14, 18}}, {{1, 6, 9, 10, 14, 17}, {2, 5, 7, 12, 15, 16}, {3, 4, 8, 11, 13, 18}}, {{1, 6, 9, 10, 14, 17}, {2, 5, 8, 11, 13, 18}, {3, 4, 7, 12, 15, 16}}, {{1, 6, 9, 10, 15, 16}, {2, 5, 7, 12, 14, 17}, {3, 4, 8, 11, 13, 18}}, {{1, 6, 9, 10, 15, 16}, {2, 5, 8, 11, 13, 18}, {3, 4, 7, 12, 14, 17}}}
Smaller than six sides won't work, because the d^3 possible outcomes from rolling three d-sided dice must be divisible by 6.
The linked article claims that you can't do four players with six-sided dice (even though 24 | 6^4), and is asking for the number of solutions with 12-sided ones.
--Michael
On Tue, Sep 18, 2012 at 7:25 AM, Alex Bellos < alexanderbellos@googlemail.com
wrote:
Anyone able to work out how many unique solutions there are to the Go First dice problem?
http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/sep/18...
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
On 9/20/2012 9:56 AM, Michael Kleber wrote:
On Thu, Sep 20, 2012 at 11:30 AM, Allan Wechsler<acwacw@gmail.com> wrote:
If we relax the constraint that all the dice have the same number of sides, then the two-person problem can be solved with only three "faces". One person "rolls" a device that always yields 2; the other has a coin with 1 and 3 on its faces. Does the three-person problem really need 18 faces?
Great question. If it could be done with all dice having #faces in {1,2,3,6}, then it could be expressed as a 3d6 solution (with appropriately-clustered numbers), and if my computation of the eleven 3d6 solutions is right, then none of them fit the bill. But this doesn't rule out e.g. something using a d4.
--Michael
Or just one die with the faces labelled by the six permutations of Al, Bob, and Charlie (or whoever the three players are). Brent
participants (5)
-
Alex Bellos -
Allan Wechsler -
Fred lunnon -
meekerdb -
Michael Kleber